Scottish Highers Maths Exponential and Logs: The graphs of $y = x^2 + 2x + 3$

The graphs of $y = x^2 + 2x + 3$ and $y = 2x^2 + x + 1$ are shown below - Scottish Highers Maths - Question 8 - 2022

Question 8

The graphs of $y = x^2 + 2x + 3$ and $y = 2x^2 + x + 1$ are shown below. The graphs intersect at the points where $x = -1$ and $x = 2$. (a) Express the shaded area... show full transcript

Worked Solution & Example Answer:The graphs of $y = x^2 + 2x + 3$ and $y = 2x^2 + x + 1$ are shown below - Scottish Highers Maths - Question 8 - 2022

Step 1

Express the shaded area, enclosed between the curves, as an integral.

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Answer

To express the shaded area between the two curves as an integral, we first identify the upper and lower functions in the interval of interest. The area $A$ can be represented by the following integral:

$A = \int_{-1}^{2} ((2x^2 + x + 1) - (x^2 + 2x + 3)) \, dx$

Simplifying the integrand:

$A = \int_{-1}^{2} (x^2 - x - 2) \, dx$

Step 2

Evaluate the shaded area.

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Answer

To evaluate the area, we first compute the integral:

$A = \int_{-1}^{2} (x^2 - x - 2) \, dx$

Calculating the integral:

$A = \left[ \frac{x^3}{3} - \frac{x^2}{2} - 2x \right]_{-1}^{2}$

Evaluating this at the bounds:

At $x = 2$ : $A(2) = \frac{2^3}{3} - \frac{2^2}{2} - 2(2) = \frac{8}{3} - 2 - 4 = \frac{8}{3} - \frac{6}{3} - \frac{12}{3} = \frac{8 - 6 - 12}{3} = \frac{-10}{3}$
At $x = -1$ : $A(-1) = \frac{(-1)^3}{3} - \frac{(-1)^2}{2} - 2(-1) = -\frac{1}{3} - \frac{1}{2} + 2 = -\frac{1}{3} - \frac{3}{6} + \frac{12}{6} = -\frac{1}{3} + \frac{9}{6} = -\frac{1}{3} + \frac{3}{2} = \frac{3 imes 2 - 2}{6} = \frac{6 - 2}{6} = \frac{4}{6} = \frac{2}{3}$

Now, we subtract:

$A = \frac{-10}{3} - \left(-\frac{2}{3}\right) = -\frac{10}{3} + \frac{2}{3} = -\frac{8}{3}$

Therefore, the area is:

$\text{Area} = \frac{8}{3} ext{ square units}.$

The graphs of $y = x^2 + 2x + 3$ and $y = 2x^2 + x + 1$ are shown below - Scottish Highers Maths - Question 8 - 2022

Worked Solution & Example Answer:The graphs of $y = x^2 + 2x + 3$ and $y = 2x^2 + x + 1$ are shown below - Scottish Highers Maths - Question 8 - 2022

Express the shaded area, enclosed between the curves, as an integral.

Evaluate the shaded area.

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